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Quick C1 equation Q

Would 3^(x+2) be y * 3^2 = 9y
and 3^(1-x) be -y * 3 = -3y

therefore 9y - 3y = 12 => 6y = 12 => y=2 => 3^x=2

or have I got this wrong
(edited 13 years ago)
Reply 1
Avatar for Ben121
Ben121
4
You know that xa∗xb=xa+bx^a * x^b = x^{a+b} and xa/xb=xa−bx^a / x^b = x^{a-b}

How can you use that to break up 3x+23^{x+2} and 31−x3^{1-x}

Your first line is correct, your second not so.
Reply 2
Avatar for Expendable
Expendable
OP
11
Original post by Ben121
You know that xa∗xb=xa+bx^a * x^b = x^{a+b} and xa/xb=xa−bx^a / x^b = x^{a-b}

How can you use that to break up 3x+23^{x+2} and 31−x3^{1-x}

Your first line is correct, your second not so.


3^(1-x) = 3/y = 3y^(-1)

therefore for b) 9y + 3y^-1 = 12
which solves down to (9y-3)(y-1)=0 (after multiplying by y, re-organizing it and factorising), which solves to give 1/3 or 1 as answer?
Reply 3
Avatar for Ben121
Ben121
4
That gives you y = 1/3 or y = 1.

Substitute that into y = 3^x and solve for x.
Reply 4
Avatar for Expendable
Expendable
OP
11
Original post by Ben121
That gives you y = 1/3 or y = 1.

Substitute that into y = 3^x and solve for x.


1 = 3^x, therefore x=0
1/3 = 3^x, therefore x=-1

right? thankyou :biggrin:
Reply 5
Avatar for Ben121
Ben121
4
Correct

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